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applied cryptography - protocols, algorithms, and source code in c

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Unformatted text preview: t size took a Cray computer just a few hours [440]. In 1988, Carl Pomerance designed a modular factoring machine, using custom VLSI chips [1259]. The size of the number you would be able to factor depends on how large a machine you can afford to build. He never built it. In 1993, a 120-digit hard number was factored using the quadratic sieve; the calculation took 825 mips-years and was completed in three months real time [463]. Other results are [504]. Today’s factoring attempts use computer networks [302, 955]. In factoring a 116-digit number, Arjen Lenstra and Mark Manasse used 400 mips-years—the spare time on an array of computers around the world for a few months. In March 1994, a 129-digit (428-bit) number was factored using the double large prime variation of the multiple polynomial QS [66] by a team of mathematicians led by Lenstra. Volunteers on the Internet carried out the computation: 600 people and 1600 machines over the course of eight months, probably the largest ad hoc multiprocessor ever assembled. The calculation was the equivalent of 4000 to 6000 mips-years. The machines communicated via electronic mail, sending their individual results to a central repository where the final steps of analysis took place. This computation used the QS and five-year-old theory; it would have taken one-tenth the time using the NFS [949]. According to [66]: “We conclude that commonly used 512-bit RSA moduli are vulnerable to any organization prepared to spend a few million dollars and to wait a few months.” They estimate that factoring a 512-bit number would be 100 times harder using the same technology, and only 10 times harder using the NFS and current technology [949]. To keep up on the state of the art of factoring, RSA Data Security, Inc. set up the RSA Factoring Challenge in March 1991 [532]. The challenge consists of a list of hard numbers, each the product of two primes of roughly equal size. Each prime was chosen to be congruent to 2 modulo 3. There are 42 numbers in the challenge, one ea...
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